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Stells [14]
3 years ago
12

Please please help!!!

Physics
1 answer:
klio [65]3 years ago
7 0

Answer:

The discovery of the natural radioactive decay of uranium in ... Uranium- 238, Lead-206, 4.5 billion years ... of the isotope carbon-14, which has a half-life of 5,730 years.

<h2>llHappiest♡Writerll</h2>

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If the power rating on a motor is 10W and 500J of energy is used to move a 20N object, how long is the motor used for?
Mkey [24]

The power rating on the motor is the maximum power it's ABLE to deliver. It can run with LESS power output than the rating, but if you try to run it with MORE than it's rated, the motor will overheat and eventually burn out.

" 10 watts " means " 10 Joules of enrgy per second ".

If the motor is operated at its full maximum rated capacity, then

(500 joules) / (10 joules/sec) = <em>50 seconds</em>

4 0
4 years ago
If sound waves require a medium (material) to vibrate in order to produce a sound, can sound waves travel in outer space? Why or
soldier1979 [14.2K]

Sound waves requires a medium (material) to vibrate in order to produce a sound. So sound waves cannot travel in outer space as it is vacuum.

8 0
3 years ago
What is the magnetic force (in newtons) on a particle traveling in a 1.5 T magnetic field if q = 7.5 microcoulombs and v = 1.75
Alla [95]

Given:

B(Magnetic field): 1.5 T                                                                                

q=  7.5 microcoulombs                                                                                

v= 1.75 x 10 ∧6 m/s                                                                                    

The angle ∅ between B and v is 45 °.                                                    

Now we know that F= qvB sin ∅

Substituting these values we get:

F= 7.5 x 10∧-6 x 1.75 x 10∧6 x 1.5 x sin 45                                                

 F= 16.752 N



3 0
3 years ago
A 5.4-kg ball is thrown into the air with an initial velocity of 35.2 m/s.a
kompoz [17]
The answer is

Ekin = 1/2 * m * v^2
Ekin = 1/2 * 5,4 * 35,2^2
Ekin = 3345,41 j (joule)
4 0
3 years ago
How many excess electrons must be present on each sphere if the magnitude of the force of repulsion between them is 4.57×10−21 n
hichkok12 [17]

Answer:

891 excess electrons must be present on each sphere

Explanation:

One Charge = q1 = q

Force = F = 4.57*10^-21 N  

Other charge = q2 =q

Distance = r = 20 cm = 0.2 m  

permittivity of free space = eo =8.854×10−12 C^2/ (N.m^2)  

Using Coulomb's law,

F=[1/4pieo]q1q2/r^2

F = [1/4pieo]q^2 / r^2

q^2 =F [4pieo]r^2

q =  r*sq rt F[4pieo]

 q=0.2* sq rt[ 4.57 x 10^-21]*[4*3.1416*8.854*10^-12]

q = 1.42614*10^ -16 C

number of electrons = n = q/e=1.42614*10^ -16 /1.6*10^-19

n =891

 891 excess electrons must be present on each sphere  

5 0
3 years ago
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